See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: When 4-aminophenol reacts with acetic anhydride (Ac2O), both the -OH and -NH2 groups can potentially be acetylated. The key question is which group reacts preferentially under the given conditions. Step 1: Identify the nucleophilic groups in 4-aminophenol. There are two nucleophilic groups: the phenolic -OH and the aromatic -NH2. Step 2: Compare nucleophilicity. The -NH2 group (amine) is a stronger nucleophile than the -OH group (phenol) because nitrogen is less electronegative than oxygen, making the lone pair on nitrogen more available for nucleophilic attack. Step 3: Reaction with Ac2O. At 100°C with one equivalent of Ac2O, the more nucleophilic -NH2 group reacts preferentially with acetic anhydride to form an amide bond (-NHAc), leaving the -OH group unreacted. Step 4: Product identification. The product retains the -OH group and has the -NH2 converted to -NHC(O)CH3 (acetylamino group). This gives N-(4-hydroxyphenyl)acetamide, commonly known as paracetamol (acetaminophen), which is option (c): a benzene ring with -OH at one position and -NHAc at the para position. Why other options fail: - Option (a): Both -OH and -NH2 are acetylated, which would require two equivalents of Ac2O and more forcing conditions. - Option (b): Only the -OH is acetylated, leaving -NH2 free. This would not occur preferentially since -NH2 is more nucleophilic. - Option (d): Shows OAc on the ring and NHAc at a meta position, and also shows a non-aromatic ring, which is inconsistent with the starting material (para-substituted benzene). Therefore, the correct answer is C.